Answer
$-5+12i$
Work Step by Step
Note: the standard form of a complex number is $a+bi$ where $i = \sqrt -1$ and $a$ and $b$ are real numbers.
Note: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(2+3i)^{2}$
The complex number can be rewritten as:
$(2+3i)^{2} = (2+3i)(2+3i)$
Expand out the complex number:
$(2+3i)^{2} = 4+6i+6i+9i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(2+3i)^{2} = 4+6i+6i+9(-1)$
Simplify:
$(2+3i)^{2} = 4+6i+6i-9$
Combine like terms, to express the complex number in standard form:
$-5+12i$
